Find roots of polynomial $$ p(x) = 2s^3+3s^2+6s+\frac{3125}{100} $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -2.6249\\[1 em]s_2 &= 0.5624+2.3741i\\[1 em]s_3 &= 0.5624-2.3741i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} 2s^3+3s^2+6s+\frac{3125}{100} & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 200s^3+300s^2+600s+3125 & = 0 \end{aligned} $$Step 2:
Polynomial $ 200s^3+300s^2+600s+3125 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.
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